Question

Q2. Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH (K 1.8 x10)

Answer 1

Sol Given 4s ml of oov m NavH is added sv ml uf o loom CH3 CooH Ka 8x lo pH- 2 Explanation: of resulting sol uwitl defend on this Deutaliuhon Yeuehion is Complete or not. Tf to complete neutralizuhion reaahon pHof sofM is not euu to 7. ff neutekze weak acid comPletehy . the pH will be conjugute buse, CHcooH is weak aud wile Nao is strvng base nic e of reauhon is OHtCH3ooH C3Coo H Cun -How many moles one adding he reuhon is cakutate hy- ca Concentrelhun Va Volume C2 nacid EOioo50 moles of CH3looH 0100MK45103L 4smLASxi2) buse E 0o045 moles of NaoH The moles of oi ions Completely Consumbed by rea? 2. no. of moles of aueic.memuin in sol? nauid O.0OSD- o'00k5 0.o00So moles

To calculate p by uing the formulu buse Conjugde -) - Pka t Jog weeak auid The new Contentrahion of aid is find by totıl volume of soln V Vacid +Vpuse. = S0 ml + 45 m = 45ml.C:gL The concentrshon is o 00050 moles (CHcooн] 3 OOo4S moles IS XIO3L O047 m pPat dog o'047 m But. Ka is given .. Ka > 18X{63 PKa =og ka .. РКа - 474 o 047M 4.74 og OOoS2M eH 70.3560 PH 5.6g